Refer to Example 5 and test the hypothesis H0: F = G against the alternatives HA'':F≠G

Question

Refer to Example 5 and test the hypothesis [latex]H_{0}\colon F=G[/latex] against the alternatives [latex]H_{A}^{\prime\prime}\colon F eq G[/latex] and [latex]H_{A}\colon F\gt G.[/latex]

Accepted Answer

In Example 5 we saw that [latex]R_{X}=26\ (\mathrm{and~}R_{Y}=19)[/latex] Since m = 5 and n = 4, relation (20) gives: U = 11.
[latex]U=R_{X}-\frac{m(m+1)}{2}=m n+\frac{n(n+1)}{2}-R_{Y}.~~~~~~~~~~~~~~~~(20)[/latex]
From the tables cited above, we have: P(U ≤ 2) = P(U ≥ 17) = 0.032. So, [latex]C_{1}=2,\ C_{2}=17,\operatorname{and}\ H_{0}[/latex] is rejected in favor of [latex]H_{A}^{\prime\prime}[/latex] at the level of significance 0.064. From the same tables, we have that P(U ≤ 3) = 0.056, so that C = 3, and [latex] extstyle H_{0}[/latex] is rejected in favor of [latex] extstyle H_{A}[/latex] at level of significance 0.056. The results stated in (23) are formulated as a lemma below.
[latex]E U={\frac{m n}{2}},\quad V\! a r(U)={\frac{m n(m+n+1)}{12}}.\qquad\qquad\qquad(23)[/latex]

LEMMA 2

With U defined by (19), the relations in (23) hold true under [latex]H_0[/latex].

[latex]U=\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n}u(X_{i}-Y_{j}).\qquad\qquad\qquad\qquad\qquad(19)[/latex]

LEMMA 3
With U, EU, and Var(U) defined, respectively, by (19) and (23),

[latex]{\frac{U-E U}{s.d.(U)}}\xrightarrow[m,n o \infty]{d} \,Z\sim{{N}}(0,1).\qquad\qquad\qquad\qquad(24)[/latex]

Question 15.6: Refer to Example 5 and test the hypothesis H0: F = G against......

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